The solution of the differential equation xdy/dx = 2x^2 + 1; x ≠ 0 with y = 1 when x = 1, is Option 1: xy = _e|x| Option 2: y = x^3 + 2x^2 Option 3: y = x^2 - _e|x| Option 4: y = x^2 + _e|x|

The answer is 'y = x^2 + _e|x|'. View full solution:

CUET Mathematics - The solution of the differential equation xdy/dx = 2x^2 + 1; x ≠ 0 with y = 1 when x = 1, is | PYQs + Solutions

CUET Mathematics 2026 18th May Shift 1

Calculus

Easy

common

$xy = \log_e|x|$

$y = x^3 + 2x^2$

$y = x^2 - \log_e|x|$

$y = x^2 + \log_e|x|$

✅ Correct Option: 4

Related questions:

15th May Shift 1

If $c$ is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I (Differential Equation)		LIST-II (Solution)
A.	$\dfrac{dy}{dx} = 1 + x + y + xy$	I.	$\log_e \vert y \vert = 2x + \log_e (x-1)^2 + c$
B.	$(x-1)\dfrac{dy}{dx} = 2xy$	II.	$y = cx$
C.	$\dfrac{dy}{dx} = 1 - x + y - xy$	III.	$\log_e \vert 1+y \vert = x + \dfrac{x^2}{2} + c$
D.	$x\,dy = y\,dx$	IV.	$\log_e \vert 1+y \vert = x - \dfrac{x^2}{2} + c$

Choose the correct answer from the options given below:

18th May Shift 2

The general solution of the differential equation $\tan\left(\dfrac{(1+x^2)y'+2xy}{2x}\right)=1+x^2$ , is

(Consider $\log_e x=\log x$ )

(where C is an arbitrary constant)

12th May Shift 2

Which of the following differential equations has $y=x$ as one of the particular solution?